Unformatted text preview: rantees there is an
such that a solution
exists and is unique for
. The maximal interval of existence is the largest
interval of time that includes for which a solution to [1] exists.
Example. The initial value problem
.
is locally Lipschitz on
initial condition to find and Separate variables and apply the .
The maximal interval of existence is . Note that Theorem 3.17 (Maximal Interval of Existence). Let
Lipschitz. Then there is a maximal open interval
value problem [1] has a unique solution
.
Proof. Let is not in the domain of . ■ be open and
containing be locally
such that the initial . Theorem 3.10 guarantees there is a ball
, a time interval
and a unique solution
of [1] such that
. Let
Then
and we can apply theorem 3.10 again. Let
Then there is a ball
, a time interval
and a unique solution
such that . .
is not empty and uniqueness implies
on
Continue this
construction both backwards and forwards in time. Note that every endpoint
lies in .
The maximal interval of existence is the union of all such intervals. Let
be the unique
solution constructed. 17
Chapter 3
Suppose that has a closed endpoint, for example
. Then
and the solution
could be extended further, contradicting the claim that is maximal. Therefore is open. □
is open. ■ Example (continued). Notice that Theorem 3.18 (Escape from any compact set). Suppose is an open set and
is
locally Lipschitz. Let
be the maximal interval of existence for [1]. If is finite, then
for any compact set
there is a
such that
. Similarly, if is finite, then
for any compact
there is a
such that
.
Idea of Proof. If is finite and
, and extend using theorem 3.10.
sketch , , , then we can define , with , extension □ , Example (continued) sketch
,
solution escapes from any compact set in
or as Corollary 3.19. If does not exist or is finite then either , both branches of
or as
.■ The Proof. If the limit exists it cannot be in , since then the solution could be extended. However,
every point
for
. Therefore
.□
Example (continued). sketch ,  indicate or and the flow .■
Example. The initial value problem
.
is locally Lipschitz on .
is the solution corresponding to
separate variables and apply the initial condition to find Assume .
For
sketch
interval of existence is ,
and , vertical asymptote,
does not exist. ■ . The maximal ,...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .
 Fall '14
 MarcEvans

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